Q.

Uniform cylinder of density ρ  and cross -sectional area A floats in equilibrium in two non- mixing liquids of densities ρ1andρ2 as shown. Length of the part of cylinder in air is 'h' & lengths of part of cylinder immersed in the liquid are h1&h2 as shown. Question Image                            

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a

h=h1[ρ1ρ+1]h2[ρ2ρ+1]

b

h=h1[ρ1ρ1]+h2[ρ2ρ1]

c

The cylinder is depressed in such a way that its top surface is just covered by the liquid of density ρ1 and then released. The restoring farce acting on cylinder isF=hΔρ2g

d

In choice (c) above, acceleration of cylinder is a=h(ρ1+ρ2)g(h+h1+h2)ρ

answer is C, B.

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Detailed Solution

weight =Fb

mg=Fb  due to liquid (1) +Fb  due to liquid (2)

vρg=v1ρ1g+v2ρ2g

vρ=v1ρ1+v2ρ2

Ah1ρ=Ah1ρ1+Ah2ρ2

h1ρ=h1ρ1+h2ρ2

(h+h1+h2)ρ=h1ρ1+h2ρ2

hρ+h1ρ+h2ρ2=h1ρ1+h2ρ2

h1ρ=h1(ρ1ρ)+h2(ρ2ρ)

h=h(ρ1ρρ)+h2(ρ2ρρ)

h=h1(ρ1ρ1)+h2(ρ2ρ1)

When cylinder is depressed in such a way that its top surface is just covered by the liquid of densityρ1  and then released its height goes in liquid (2). So extra buoyant force due to liquid (2) acts on it. Due to this restoring force, cylinder is accelerated upward.

*Restoring force acting on the cylinder

F=hAρ2g

=hAρ2g

*Acceleration of the cylinder

a=Fm=hAρ2g(h+h1+h2)

=hρ2g(h+h1+h2)ρ

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